VECTORS

[sonal]

A boat sails 6 km West ,then 5 km Northwest.Use trigonometry to find the boat's distance and bearing from its starting point.

Please solve this thanks

 

[pka]

A boat sails 6 km West ,then 5 km Northwest.Use trigonometry to find the boat's distance and bearing from its starting point.
Please solve this thanks

Question, why should we solve this? Isn't it assigned to you to solve?
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[sonal]

I have made multiple attempts to solve but am still stuck, so far have figured out the direction bearing which would be 309.9 degrees as tan inverse of 5/6 is 39.8. since the bearing is measured clockwise from true north it 270 + 39.8 degrees is 309.9. am not sure how to figure out the distance though.

 

[MarkFL]

Draw a diagram, and you will find you have two sides of a triangle, whose lengths you know, and you know the angle subtending them...sounds like a job for the Law of Cosines (to get the length of the third unknown side).

 

[sonal]

How would i use cosine law as i don't have the angle opposite to the unknown side, with the cosine law i would still have two unknown unless there is a way to figure out the value of the obtuse angle that is opposite the unknown length/distance

 

[MarkFL]

How would i use cosine law as i don't have the angle opposite to the unknown side, with the cosine law i would still have two unknown unless there is a way to figure out the value of the obtuse angle that is opposite the unknown length/distance

The ship first heads due west, and then changes directions to northwest, which means the two sides are subtended by the angle:

[MATH]\theta=(90+45)^{\circ}=135^{\circ}[/MATH]

 

[sonal]

Oh ok I understand thankyou , also is my bearing (309.9 degrees) correct

 

[MarkFL]

Oh ok I understand thankyou , also is my bearing (309.9 degrees) correct

Let's see:

[MATH]s_x=5\cos(180^{\circ})+6\cos(135^{\circ})=-(5+3\sqrt{2})[/MATH]
[MATH]s_y=5\sin(180^{\circ})+6\sin(135^{\circ})=3\sqrt{2}[/MATH]
[MATH]\beta=\frac{180^{\circ}}{\pi}\left(\frac{3\pi}{2}-\arctan\left(\frac{s_y}{s_x}\right)\right)\approx294.656498287^{\circ}[/MATH]
I get a different bearing than you did.